4 ms·

Could you provide an example? I'm not seeing it. Is there some integer that is not also a rational number?

klodolph 2y ago
How about this example: the natural number 0 is not a member of the integers, from a set theoretic perspective. In the natural numbers, 0 = {} However, {} is not an element of the integers. This is not something I expect to be easy to understand. This is the standard set theoretic definition of integers that mainstream mathematicians use. This is not some esoteric, fringe theory.
- Affric 2y ago
  Isn’t the empty set a subset of all sets? A quick look at Wikipedia indicates that there exist other constructions.
  - klodolph 2y ago
    Yes, the empty set is a subset of all sets. I think some wires got crossed somewhere because whether zero is a subset of some other set doesn’t fit into the questions we are trying to answer. You could pick a construction where the natural numbers are a subset of the integers. This is trivial, but this is a poor strategy overall, because you can always find a bigger set of numbers to work with. You can’t take the “biggest” set of numbers and then define all other sets of numbers as subsets of that. It would be kind of like trying to count down from infinity.
    - Affric 2y ago
      I see where I have gone wrong. Element vs subset. Classic error when thinking fast. Thank you for the thought provoking comments.
    - SAI_Peregrinus 2y ago
      The Surreal Numbers manage to be a sort of "biggest" ordered field, in NBG, though they're a proper class not a set. All other ordered fields are subfields of the Surreals. Of course "ordered" is doing a lot, since it excludes the complex numbers, vectors, bivectors, etc. Whether elements of some object that isn't an ordered field should be considered "nubmers" is related but different question.
  - enugu 2y ago
    Yes, {} is a subset(⊂) of all sets, but it is not a member(∊) of all sets. For instance, {} is not a member of {{{}}}. In the Von-Neumann definition, 1:={0}={{}}. So, 0∊1, but 0 is not a member of {1} which is the above set. https://en.wikipedia.org/wiki/Set-theoretic_definition_of_natural_numbers https://en.wikipedia.org/wiki/Set-theoretic_definition_of_na...
- interroboink 2y ago
  Ah, thank you. It does actually make some sense, having looked into it some more now (: It looks like the approach of defining ℤ in terms of ℕ is much more tedious to deal with overall, so can see the advantages.
Cu3PO42 2y ago
Conceptually, yes, every integer is also a rational number. But they are represented as different objects. It's a bit like saying every int32 is also a double. Yes, every value of int32 fits into a double, but the bit pattern is different. The canonical construction for rational numbers is pairs (a, b) which we interpret as a/b. An integer k "is the same as" (k, 1). So it might be more correct to say every integer has the same value as some rational number. Of course this distinction is pointless most of the time, so we don't worry about it. This is not unique to integers and rationals. It also applies to naturals and integers, rationals and reals, etc.
- pyrolistical 2y ago
  But an int32 1 isn’t the same as a mathematic object 1. So of course your analogy doesn’t work
  - klodolph 2y ago
    By definition, analogies are between things that aren’t the “same thing”.
  - Cu3PO42 2y ago
    And a double 1.0 isn't the same as a rational 1/1. But I also didn't claim either of these things.
- _nalply 2y ago
  I understand that this might be a problem in a programming language, even if you decide to ignore that mathematical sets have infinitively many elements. When I programmed an RPN calculator supporting quotients and complex numbers, I had to grapple with the fact that 1 (one) is a quotient (with denominator 1) and a complex number (with imaginary part 0). I solved it by defining the functions and operators taking a certain type and coercing the arguments to that type. One example, the logarithm can take complex numbers. If I had a quotient argument, I coerce it to a complex number before passing it to the logarithm. However I assumed that this is only a problem in programming languages. I am a bit surprised that mathematics also seems to be affected. I am going to study it a bit.
deleted 2y ago
[deleted]

How about this example: the natural number 0 is not a member of the integers, from a set theoretic perspective. In the natural numbers, 0 = {} However, {} is not an element of the integers. This is not something I expect to be easy to understand. This is the standard set theoretic definition of integers that mainstream mathematicians use. This is not some esoteric, fringe theory.

Affric 2y ago
Isn’t the empty set a subset of all sets? A quick look at Wikipedia indicates that there exist other constructions.
- klodolph 2y ago
  Yes, the empty set is a subset of all sets. I think some wires got crossed somewhere because whether zero is a subset of some other set doesn’t fit into the questions we are trying to answer. You could pick a construction where the natural numbers are a subset of the integers. This is trivial, but this is a poor strategy overall, because you can always find a bigger set of numbers to work with. You can’t take the “biggest” set of numbers and then define all other sets of numbers as subsets of that. It would be kind of like trying to count down from infinity.
  - Affric 2y ago
    I see where I have gone wrong. Element vs subset. Classic error when thinking fast. Thank you for the thought provoking comments.
  - SAI_Peregrinus 2y ago
    The Surreal Numbers manage to be a sort of "biggest" ordered field, in NBG, though they're a proper class not a set. All other ordered fields are subfields of the Surreals. Of course "ordered" is doing a lot, since it excludes the complex numbers, vectors, bivectors, etc. Whether elements of some object that isn't an ordered field should be considered "nubmers" is related but different question.
- enugu 2y ago
  Yes, {} is a subset(⊂) of all sets, but it is not a member(∊) of all sets. For instance, {} is not a member of {{{}}}. In the Von-Neumann definition, 1:={0}={{}}. So, 0∊1, but 0 is not a member of {1} which is the above set. https://en.wikipedia.org/wiki/Set-theoretic_definition_of_natural_numbers https://en.wikipedia.org/wiki/Set-theoretic_definition_of_na...
interroboink 2y ago
Ah, thank you. It does actually make some sense, having looked into it some more now (: It looks like the approach of defining ℤ in terms of ℕ is much more tedious to deal with overall, so can see the advantages.

Cu3PO42 2y ago

Conceptually, yes, every integer is also a rational number. But they are represented as different objects. It's a bit like saying every int32 is also a double. Yes, every value of int32 fits into a double, but the bit pattern is different. The canonical construction for rational numbers is pairs (a, b) which we interpret as a/b. An integer k "is the same as" (k, 1). So it might be more correct to say every integer has the same value as some rational number. Of course this distinction is pointless most of the time, so we don't worry about it. This is not unique to integers and rationals. It also applies to naturals and integers, rationals and reals, etc.

pyrolistical 2y ago
But an int32 1 isn’t the same as a mathematic object 1. So of course your analogy doesn’t work
- klodolph 2y ago
  By definition, analogies are between things that aren’t the “same thing”.
- Cu3PO42 2y ago
  And a double 1.0 isn't the same as a rational 1/1. But I also didn't claim either of these things.
_nalply 2y ago
I understand that this might be a problem in a programming language, even if you decide to ignore that mathematical sets have infinitively many elements. When I programmed an RPN calculator supporting quotients and complex numbers, I had to grapple with the fact that 1 (one) is a quotient (with denominator 1) and a complex number (with imaginary part 0). I solved it by defining the functions and operators taking a certain type and coercing the arguments to that type. One example, the logarithm can take complex numbers. If I had a quotient argument, I coerce it to a complex number before passing it to the logarithm. However I assumed that this is only a problem in programming languages. I am a bit surprised that mathematics also seems to be affected. I am going to study it a bit.

deleted 2y ago

[deleted]